Given:
The equation is,
[tex]2\cos ^2x+9\sin x=3\sin ^2x[/tex]
Explanation:
Simplify the equation by using trigonometric identity.
[tex]\begin{gathered} 2(1-\sin ^2x)+9\sin x=3\sin ^2x \\ 2-2\sin ^2x+9\sin x=3\sin ^2x \\ 5\sin ^2-9\sin x-2=0 \end{gathered}[/tex]
Assume sin x = t, then
[tex]5t^2-9t-2=0[/tex]
Solve the equation by splitting the middle term.
[tex]\begin{gathered} 5t^2-10t+t-2=0 \\ 5t(t-2)+1(t-2)=0 \\ (5t+1)(t-2)=0 \\ t=-\frac{1}{5},2 \end{gathered}[/tex]
So,
[tex]\sin x=-\frac{1}{5}\text{ or sin x = 2}[/tex]
There is no possible value of x, for sin x = 2.
Determine the value of x by using sin x = -1/5.
[tex]\begin{gathered} \sin x=-\frac{1}{5} \\ x=\sin ^{-1}(-\frac{1}{5}) \\ \approx-0.2014,-2.9402 \end{gathered}[/tex]
So possible values of x are -0.2014 and -2.9402.
